Isoelectronic Ions: Rank Ionic Radii With Worked Practice

A fluoride ion, F⁻, and a magnesium ion, Mg²⁺, each have ten electrons. Mg²⁺ is still smaller. Their electron configurations match, but their nuclei don't: fluorine has nine protons, magnesium has twelve.

That distinction gives you a method for ranking isoelectronic ions. Check the electron configuration first, then compare proton counts. The practice worksheet below includes a mixed set, so part of the task is deciding which ions you can rank together.

These exercises use monatomic ground-state ions with matching closed-shell configurations. The answers are qualitative size orders, without numerical radii or molecular comparisons.

Start with the electron count

For atoms and monatomic ions, isoelectronic species have the same electron configuration. Count electrons first, then check their arrangement. All the ions in this worksheet have one of these two ground-state configurations:

Electrons	Closed-shell configuration	Shorthand
10	1s² 2s² 2p⁶	[Ne]
18	1s² 2s² 2p⁶ 3s² 3p⁶	[Ar]

The atomic number, Z, gives the proton count. Ion formation changes the number of electrons while leaving Z unchanged:

electrons = Z − signed charge

For Mg²⁺, subtract +2: 12 − (+2) = 10. For F⁻, subtract −1: 9 − (−1) = 10. Positive charge means electrons were lost; negative charge means electrons were gained. Keeping the sign in the calculation helps prevent a 2− ion from accidentally losing two electrons on paper.

Our electron configuration practice lesson explains how to derive configurations, including which subshells lose electrons when ions form. Use it if the electron total is clear but the configuration isn't. A matching total alone isn't a substitute for checking the ground-state arrangement in an unfamiliar problem.

You can look up Z in our periodic table reference. Don't use a mass number or decimal atomic weight for these counts.

Within one series, more protons means a smaller radius

In an isoelectronic series, greater nuclear charge means a smaller radius. The stronger attraction draws the electron cloud closer to the nucleus. OpenStax's ionic-radius discussion describes this relationship.

Matching configurations tell you which subshells the electrons occupy. They don't fix how far the cloud extends. F⁻ and Mg²⁺ can both have the [Ne] configuration and different radii.

For a written solution, leave enough working to check your method:

1. Calculate each electron count.
2. Confirm the configurations and separate the ions into isoelectronic series.
3. Compare proton counts within each series.
4. Write the radius order the question requests.

Here, < means smaller radius than and > means larger radius than. Label the ends “smallest” and “largest” before sorting if you tend to reverse inequalities.

Worked example: smallest to largest

Arrange F⁻, Al³⁺, and Mg²⁺ in increasing ionic radius. Their atomic numbers are 9, 13, and 12.

Ion	Electron calculation	Electrons	Protons	Configuration
F⁻	9 − (−1)	10	9	[Ne]
Al³⁺	13 − (+3)	10	13	[Ne]
Mg²⁺	12 − (+2)	10	12	[Ne]

All three belong to one series. Their proton counts decrease in the order 13 > 12 > 9, so their radii increase in that order:

Al³⁺ < Mg²⁺ < F⁻

An explanation needs both parts: “All three ions have the [Ne] configuration. Al³⁺ has the most protons and therefore the smallest radius; F⁻ has the fewest and the largest radius.”

Worked example: largest to smallest

Arrange Ca²⁺, Cl⁻, and K⁺ in decreasing ionic radius. Here, Z is 20, 17, and 19.

Ion	Electron calculation	Electrons	Protons	Configuration
Ca²⁺	20 − (+2)	18	20	[Ar]
Cl⁻	17 − (−1)	18	17	[Ar]
K⁺	19 − (+1)	18	19	[Ar]

The largest ion goes first, so start with the lowest proton count:

Cl⁻ > K⁺ > Ca²⁺

Check an adjacent pair. K⁺ and Ca²⁺ each have eighteen electrons, but Ca²⁺ has one more proton. It belongs at the smaller-radius end, which agrees with the line above.

Check the series before using a shortcut

Within these series, the more positive ion also has more protons. That makes “more positive means smaller” look like a useful shortcut. Across unrelated ions, the charge label alone doesn't establish a radius order. Check the configurations first.

The same limit applies to “more protons means smaller.” Mg²⁺ has ten electrons, while Ca²⁺ has eighteen. Their configurations differ, so comparing their proton counts alone doesn't justify a size order. Some mixed pairs can be compared using another periodic trend; the isoelectronic rule alone doesn't rank an arbitrary mixed set.

If a problem asks for numerical radii, check what its table measures. Shannon's original ionic-radius paper distinguishes coordination number—the number of surrounding neighbors—and electronic spin state, among other factors. An ion's radius isn't one context-free number. Use comparable entries rather than combining values from different environments. The worksheet here asks only for the introductory qualitative orders.

Isoelectronic ions practice worksheet

Use a blank sheet and cover the answer key. All the atomic numbers you need are supplied. For the counting table, use columns for ion, protons, electron calculation, electrons, and configuration. Add a sentence explaining each radius order.

1. Complete the table for this shuffled set: Cl⁻ (Z = 17), Na⁺ (11), S²⁻ (16), Al³⁺ (13), K⁺ (19), O²⁻ (8), Ca²⁺ (20), and Mg²⁺ (12).
2. Separate those eight ions into isoelectronic groups. Give the common configuration of each group.
3. Rank the ten-electron group from smallest to largest radius.
4. Rank the eighteen-electron group from largest to smallest radius.
5. Someone groups S²⁻ and O²⁻ together because both have a 2− charge. Does that establish an isoelectronic pair? Show the counts.
6. A counting table lists Al³⁺ as having 10 protons and 10 electrons. Which entry needs repair? Explain what the 3+ charge changed.
7. A learner writes “Ca²⁺ < Mg²⁺ because calcium has more protons.” Identify the missing check. Can the isoelectronic rule justify that line?
8. An ion has 16 protons and 18 electrons. Identify its element and charge, then place it in one of your groups. Repeat for an ion with 13 protons and 10 electrons.

Answers and explanations

1. Check the counting table

Ion	Protons	Electron calculation	Electrons	Configuration
Cl⁻	17	17 − (−1)	18	[Ar]
Na⁺	11	11 − (+1)	10	[Ne]
S²⁻	16	16 − (−2)	18	[Ar]
Al³⁺	13	13 − (+3)	10	[Ne]
K⁺	19	19 − (+1)	18	[Ar]
O²⁻	8	8 − (−2)	10	[Ne]
Ca²⁺	20	20 − (+2)	18	[Ar]
Mg²⁺	12	12 − (+2)	10	[Ne]

If a negative ion came out with fewer electrons than protons, fix the sign step before attempting the radius questions.

2. Separate the two series

The ten-electron [Ne] group contains O²⁻, Na⁺, Mg²⁺, and Al³⁺. The eighteen-electron [Ar] group contains S²⁻, Cl⁻, K⁺, and Ca²⁺.

The rule lets you sort each group. It doesn't give a single combined ionic radius order for all eight ions.

3. Increase radius through the ten-electron group

Al³⁺ < Mg²⁺ < Na⁺ < O²⁻

Their proton counts are 13, 12, 11, and 8, respectively. The highest proton count belongs at the smallest-radius end. If you wrote O²⁻ < Na⁺ < Mg²⁺ < Al³⁺, you sorted the proton counts from low to high but reversed the radius relationship.

4. Decrease radius through the eighteen-electron group

S²⁻ > Cl⁻ > K⁺ > Ca²⁺

The proton counts are 16, 17, 19, and 20. The lowest proton count goes first because this question starts with the largest radius. Read the inequalities as statements about size, not proton count.

5. Equal charges don't establish an isoelectronic pair

S²⁻ has 16 + 2 = 18 electrons; O²⁻ has 8 + 2 = 10. Their configurations are [Ar] and [Ne]. Each gained two electrons relative to its own neutral atom, but those atoms started with different electron counts.

6. Repair the proton column

Al³⁺ has 13 protons and 10 electrons. The electron entry is correct. The 3+ charge records three lost electrons. Removing three protons would change the element's identity rather than form an aluminum ion.

7. A different configuration needs a different comparison

The missing check is electron configuration. Ca²⁺ has eighteen electrons and [Ar]; Mg²⁺ has ten and [Ne]. They aren't isoelectronic, so the learner's method is invalid.

The proposed inequality is also reversed for this pair. Mg and Ca lie in the same periodic-table group, and their 2+ ions follow the general increase in radius down that group: Mg²⁺ < Ca²⁺. Ca²⁺ has an occupied third shell, whereas Mg²⁺ ends at the second. This is a separate trend, described in the OpenStax discussion linked above, rather than an application of the isoelectronic rule.

8. Work backward from charge balance

Use signed charge = protons − electrons.

With 16 protons, the element is sulfur. 16 − 18 = −2, giving S²⁻, in the eighteen-electron group. With 13 protons, the element is aluminum. 13 − 10 = +3, giving Al³⁺, in the ten-electron group.

More electrons than protons means negative charge; fewer means positive charge. The charge magnitude tells you how many electrons were gained or lost relative to the neutral atom.

Review the step you missed

Keep the worked rankings on paper. If you missed a question, choose a short review prompt for the cause of the error rather than memorizing the whole answer table.

Mistake	Front	Back
Reversed the effect of negative charge	O has Z = 8. How many electrons does O²⁻ have?	10; the ion gained two electrons.
Changed the nucleus during ion formation	How many protons remain in Al³⁺, given Z = 13?	13. The charge changes electron count.
Assumed matching charges meant matching configurations	Are O²⁻ and S²⁻ isoelectronic? Z is 8 and 16.	No: 10 electrons, [Ne], versus 18 electrons, [Ar].
Reversed the radius order	F⁻ and Mg²⁺ each have 10 electrons. Their proton counts are 9 and 12. Which is smaller?	Mg²⁺; its nucleus has more protons.
Applied the radius rule too early	What must match before you rank different elements' ions by proton count alone?	Their electron configurations: they must form one isoelectronic series.

Nibomo's periodic-trends flashcards include isoelectronic comparisons alongside atomic and ionic radius, shielding, and other trends. Use the deck for concise recall practice and this worksheet for showing the counting and grouping steps.

On your next attempt, cover both the reference and answer key. Write electron and proton counts beside every ion, then read the final line as “smaller than” or “larger than.” Changing the requested order should change where you write the smallest ion, not which ion it is.